The first observation to make is that, given a bounded sesquilinear form $B$ on $\mathcal{H}$, i.e., a form such that $|B(x,y)| \leq \|B\|\,\|x\|\,\|y\|$, there exists a unique bounded linear operator $T$ on $\mathcal{H}$ such that $B(x,y) = \langle x,Ty \rangle$ and $\|T\| = \|B\|$ by the Riesz representation theorem (see Pedersen, Analysis Now, Lemma 3.2.2, p.89 for details).
If $B$ is Hermitian, the identity
$$\langle x, Ty \rangle = B(x,y) = \overline{B(y,x)} = \overline{\langle y, Tx \rangle} = \langle Tx, y\rangle = \langle x, T^{\ast}y\rangle$$
shows that $T = T^{\ast}$, so $T$ is self-adjoint.
If $B$ is positive definite then of course $T$ must be injective, but it need not be invertible (an injective self-adjoint operator has dense range, of course).
It is not hard to show that your equivalence relation on scalar products implies that the operator $T$ is bounded and bounded away from zero, and as it is self-adjoint, this means that $T$ is invertible, see Pedersen, Proposition 3.2.6, page 90.
Conversely, if $T$ is bounded, self-adjoint and invertible, then $T$ is bounded away from zero and hence the scalar products $\langle x, Ty \rangle$ and $\langle x,y\rangle$ are equivalent.
It is plain that for a scalar product $B$ the condition
$$a \langle x,x \rangle \leq B(x,x) \leq b\langle x,x\rangle \quad \text{for all } x \in \mathcal H$$
implies that $\mathcal{H}$ is complete with respect to $B$, so we can sum up:
If a scalar product $B$ satisfies $a \|x\|^2 \leq B(x,x) \leq b \|x\|^2$ for all $x$ then $B(x,y) = \langle x, Ty \rangle$ for a unique bounded, invertible self-adjoint operator $T$. Conversely, for every bounded, invertible self-adjoint operator we get equivalent scalar products.
In particular, there is only one equivalence class of scalar products. It is clear that you can perform all this with bases as well, and, as Willie said, your reasoning is correct. However, these calculations seem more involved to me than the above considerations.
Added:
A closely related result that I should have mentioned is the Lax–Milgram theorem.
You may consider an H-valued random variable as a measurable function from $\Omega$ to $H$, where $H$ is equipped with its Borel $\sigma$-algebra. Then as you say, $L^2(\Omega; H)$ is the set of (equivalence classes of) random variables $f$ such that $E[\|f\|^2] < \infty$.
The inner product is $E[\langle f, g\rangle]$ as you guessed; this is indeed an inner product (easy exercise), and moreover it is complete, so $L^2(\Omega; H)$ is a Hilbert space itself.
Best Answer
There is not much you have to do: you are giving a vector space with an inner product. It's completion is a Hilbert space. The only issue is to show that the inner product is such.
The definition is fine in the sense that the functions $\{1_{[0,t]}:\ t\in[0,T]\}$ are linearly independent, so you can define $\langle f,g\rangle_H$ for all $f,g\in\mathcal E$. What is not so clear to me is that you get an inner product: mainly, you need to show that $\langle f,f\rangle_H=0$ implies $f=0$. This means that if $f=\sum_{j=1}^n\alpha_j\,1_{[0,t_j]}$, $$ 0=\langle f,f\rangle_H=\frac12\,\sum_{k,j}\alpha_k\alpha_j(t_k^{2H}+t_j^{2H}-|t_k-t_j|^{2h}), $$ implies $\alpha_1=\cdots=\alpha_n=0$. I don't see an immediate way to show that, and that's the crucial issue to solve in your questions. If you show the above, the rest is straightforward.
There are a couple issues with your reasoning, too. It is not true that $t^{1/2}$ and $t^H$ are equivalent, unless $H=1/2$. If you had $H<1/2$ and $t^H\leq c\,t^{1/2}$, you have $t^{1/2-H} \geq1/c$ for all $t$ close to zero, which is false. Similarly with $H>1/2$.
Also, to test that two norms are equivalent you have to do it for all elements of the space, not just some. So you cannot do it for $1_{[0,t]}$ only, you have to do it for all linear combinations.